Question

If $\alpha ,\beta$ roots of $x^2+px+1=0$ and $\gamma ,\delta$ are the roots of $x^2+qx+1=0$ then prove that $(\alpha -\gamma )(\beta -\gamma )(\alpha +\delta )(\beta +\delta )=q^2-p^2$

Answer 1

$x^2+px+1=0-----\left(1\right)$ roots $\to \alpha$ & $\beta$

$\alpha =\beta =-p$ and $\alpha \beta =1$

$x^2+qx+1=0-----\left(2\right)$ roots $\gamma$ and $\delta$

$(\alpha -\gamma )(\beta -\gamma )(\alpha +\delta )(\beta +\delta )$

$=(\alpha \beta -\beta \gamma -\alpha \gamma -{\gamma }^2)(\alpha \beta +\beta \delta +\alpha \delta +{\delta }^2)$

$=[1-\gamma (\alpha +\beta )+{\gamma }^2][1+\delta (\alpha +\beta )+{\delta }^2]$

$=[1+p\gamma +{\gamma }^2][1+p\delta +{\delta }^2]$

$=1+p\gamma +{\gamma }^2+p\delta +p^2\gamma \delta +{\gamma }^{2}p\delta +{\delta }^2+{\delta }^{2}p\gamma +{\left(\gamma \delta \right)}^2$

$=1+p(\gamma +\delta )+{\gamma }^2+p^2\left(1\right)+\gamma p+{\delta }^2+\delta p+1$

$=1-pq+p^2+p(\gamma +\delta )+{\delta }^2+1+{\gamma }^2$

$=2-pq-pq+p^2+{(\gamma +\delta )}^2-2\gamma \delta$

$=2-2pq++p^2++q^2-2$

$={(p-q)}^2$